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Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to \bigtriangleup ABC with scale factor \frac{3}{2} . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal

Solution

Steps of construction

  1. Draw a line segment BC = 6 cm
  2. Taking B and C as a centres, draw arc of radius AB= 4cm and AC=9cm
  3. Join AB and AC
  4. DABC is required triangle From B draw ray BM with acute angle \angle XBM
  5. Make 3 points B_{1},B_{2},B_{3}  on BM with equal distance
  6. Join B_{2}C  and B_{3}  draw B_{3}X\parallel B_{2}C  intersecting BC at X From point X draw XY||CA intersecting the extended line segment BA to Y Then \bigtriangleup BXY  is the required triangle whose sides are equal to\frac{3}{2}  of the \bigtriangleup ABC
     Justification :
    Here B_{3}X\parallel B_{2}C
    \therefore \frac{BC}{CX}= \frac{2}{1}
    \therefore \frac{BX}{BC}= \frac{BC+CX}{BC}= 1+\frac{1}{2}= \frac{3}{2}
    Also XY\parallel CA
    \bigtriangleup ABC\sim \bigtriangleup YBX
    \therefore \frac{YB}{AB}= \frac{YX}{AC}= \frac{BX}{BC}= \frac{3}{2}
    Here all the three angles are same but three sides are not same.
    \thereforeThe two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size.

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Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60^{\circ}. Also justify the
construction. Measure the distance between the centre of the circle and the point of intersection of tangents

Solution

 

Steps of construction

1. Draw a circle of radius OA = 4 cm with centre O
2. Produce OA to P such that  OA= AP= 4cm
3. Draw a perpendicular bisector of OP=8cm
4.Now taking A as Centre draw circle of radius AP = OA = 4 cm
5.Which intersect the circle at x and y
6.Join PX and PY
7.PX and PY is the tangent of the circle
Justification
In \bigtriangleup OAX we have
OA= OP= 4\, cm (Radius)
AX= 4\, cm (Radius of circle with centre A)
\therefore OAX is equilateral triangle
\angle OAX= 60^{\circ}
\Rightarrow \angle XAP= 120^{\circ}
In \bigtriangleup PAX we have
PA= AX= 4cm
\angle XAP= 120^{\circ}
\angle APX= 30^{\circ}
\Rightarrow APY= 30^{\circ}
Hence \angle XPY= 60^{\circ}

 

 

 

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Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ABC= 60^{\circ}.Construct a triangle similar to ABC with scale factor \frac{5}{7} . Justify the construction

Solution
Given : AB = 5 cm, BC = 6 cm

Steps of construction

  1. Draw line segment AB = 5 cm
  2. draw < ABO= 60^{\circ}  B taking as a centre draw an arc of radius BC=6cm
  3. Join AC, DABC is the required triangle
  4.  From point A draw any ray A{A}'   with acute angle \angle BA{A}'
  5.  Mark 7 points B_{1},B_{2},B_{3},B_{4},B_{5},B_{6},B_{7}  with equal distance.
  6. Join B_{7}B  and form B_{5}  draw B_{5}X\parallel B_{7}B\, \, BYmaking the angle equal From point X draw XY\parallel BC intersecting AC at Y. Then, DAMN is the required triangle whose sides are equal to \frac{5}{7}  of the corresponding sides of the \bigtriangleup ABC..

Justification: Here, B_{5X}\parallel B_{7}B  [by construction]
          \therefore \frac{AX}{XB}= \frac{5}{2}\Rightarrow \frac{XB}{AX}= \frac{2}{5}
Now \frac{AB}{AX}= \frac{AX+XB}{AX}
1+\frac{XB}{AX}= 1+\frac{2}{5}= \frac{7}{5}
Also, XY\parallel BC
\therefore \bigtriangleup AXY\sim \bigtriangleup ABC
\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}

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Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Solution

Steps of construction
1.   Draw line BC = 5 cm 
2.   Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3.   Join AB and AC DABC is required isosceles triangle
4    From B draw ray  B_{X} with an acute angle CB{B}'
  6.   draw  B_{1},B_{2},B_{3},B_{4}  at BX  with equal distance
7.   Join B_{3}C  and from B_{4}  draw line B_{4}D\parallel B_{3}C, , intersect extended segment BC at point D.
  8.   From  point D draw DE\parallel CA meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
\because B_{4}D\parallel B_{3}C
\therefore \frac{BC}{CD}= \frac{3}{1}\Rightarrow \frac{CD}{BC}= \frac{1}{3}
 Now\, \therefore \frac{BD}{BC}= \frac{BC+CD}{BC}= 1+\frac{CD}{BC}= 1+\frac{1}{3}= \frac{4}{3}
 Also\, DE\parallel CA
\therefore \bigtriangleup ABC\sim \bigtriangleup DBE
\frac{EB}{AB}= \frac{DE}{CA}= \frac{BD}{BC}= \frac{4}{3}

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Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

Solution
 

Steps of construction
1.   Draw two concentric circles with center O and radii 3 cm and 5 cm
2.   Taking any point P on outer circle, Join P and O
3.   Draw perpendicular bisector of OP let M be the mid point of OP
4.   Taking M as centre and OM as radius draw a circle which cuts inner circle at Q and R
5.   Join PQ and PR. Thus PQ and PR are required tangents 
On measuring PQ and PR we find that PQ = PR = 4 cm
Calculations
\bigtriangleup OQP, \angle OQP= 90^{\circ}
OP^{2}= OQ^{2}+PQ^{2}\left [using pythagoras theorem]
\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( PQ \right )^{2}
25-9= PQ^{2}
16= PQ^{2}
\sqrt{16}= PQ
4cm= PQ
Hence the length of both tangents is 4 cm.

 

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Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ABC= 60^{\circ}, divide it into triangles BCD and ABD
by the diagonal BD. Construct the triangle BD'C' similar to DBDC with scale factor  \frac{4}{3}. Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram?

Solution

Steps of construction
1.   Draw, A line AB = 3 cm
2  Draw a ray by making  \angle ABP= 60^{\circ}
3.   Taking  B as centre and radius equal to 5 cm. Draw an arc which cut BP at point C
4.   Again draw ray AX making  \angle {Q}'AX= 60^{\circ}
5.   With A as centre and radius equal to 5 cm draw an arc which cut AX at point D
6.   Join C and D Here ABCD is a parallelogram           
7.   Join BD , BD is a diagonal of parallelogram ABCD
8.   From B draw a ray BQ with any acute angle at point B i.e., \angle CBQ  is acute angle
9.   Locate 4 points B_{1},B_{2},B_{3},B_{4}  on BQ with equal distance.
10. Join B_{3}C  and from  B_{4},{C}'  parallel to B_{3}C  which intersect at point {C}'
 11. From point  {C}' draw line {C}'{D}'  which is parallel to CD
 12. Now draw a line segment {D}'{A}'  parallel to DA
 Note : Here {A}',{C}'  and {D}'  are the extended sides.
13.  {A}'B{C}'{D}' is a parallelogram in which {A}'{D}'= 6\cdot 5\, cm  and {A}'{B}= 4\, cm  and < {A}'B{D}'= 60^{\circ} divide it into triangles B{C}'{D}'  and {A}'{BD}'  by the diagonal {BD}'

           

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Two line segments AB and AC include an angle of 60^{\circ} where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP= \frac{3}{4}AB  and AQ= \frac{1}{4}AC . Join P and Q and measure the length PQ.

Solution
Given  :
AB = 5 cm and AC = 7 cm
AP= \frac{3}{4}AB\: \: \cdots 1
AQ= \frac{1}{4}AC\: \: \cdots 2
From equation 1
AP= \frac{3}{4}AB
AP= \frac{3}{4}\times 5= \frac{15}{4}\: \: \left [ \because AB= 5cm \right ]
P is any point on B
\therefore PB= AB-AP= 5-\frac{15}{4}= \frac{20-15}{4}= \frac{5}{4}cm
\frac{AP}{AB}= \frac{15}{4}\times \frac{4}{5}= \frac{1}{3}
AP:AB= 1:3
\thereforescale of a line segment AB is \frac{1}{3}


 Steps of construction
  1.   Draw line segment AB = 5 cm
  2.   Now draw ray AO which makes an angle i.e., < BAO= 60^{\circ}
  3.   Which A as center and radius equal to 7 cm draw an arc cutting line AO at C
  4.   Draw ray AP with acute angle BAP
  5.   Along AP make 4 points A_{1},A_{2},A_{3},A_{4}  with equal distance.
  6.   Join A_{4}B
  7.   From  A_{3} draw A_{3}P  which is parallel to A_{4}B  which meet AB at point P.
Then P is point which divides AB in ratio 3 : 1
   AP : PB = 3 : 1
  8.   Now draw ray AQ, with an acute angle CAQ.
  9.   Along AQ mark 4 points  B_{1},B_{2},B_{3},B_{4} with equal distance.
 10. Join  B_{4}C
  11. From B_{1}  draw B_{1}Q  which is parallel to B_{4}C  which meet AC at point Q.
  Then Q is point which divides AC in ratio 1 : 3
   AQ : QC = 1 : 3
  12. Finally join PQ and its measurement is 3.25 cm.

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Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre

Solution 
Given : Radius = 4cm

Steps of construction
1.  Draw a circle at radius r = 4 cm at point O. 
2.   Take a point P at a distance 6cm from point O and join PO
3.  Draw a perpendicular bisector of  line PO M is the mid-point of  PO.
4.   Taking M as centre draw another circle of radius equal to MO and it intersect the given circle at point Q and R  

5.   Now, join PQ and PR.
Then PQ and PR one the required two tangents.

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Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.Construct a triangle similar to it and of scale factor \frac{5}{3}

Solution 
Given :  BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3

Steps of construction

  1. Draw line BC = 6cm
  2. Taking B and C as centre mark arcs of length 4cm and 5cm respectively which intersecting each other at point A.
  3. Join BA and CA
  4. Draw a ray BX making acute angle with BC.
  5. Locate 5 Points on BX in equidistant as B1, B2, B3, B4, B5.
  6. Join B3 to C and draw a line through B5 parallel to B3C to intersect at BC extended at  {C}'.

       7. Draw a line through  {C}' parallel to AC intersect AB extended at {A}' .

Now  {A}'B{C}'  is required triangle.

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Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and \angle B= 90^{\circ}.Construct a triangle similar to it and of scale factor \frac{2}{3} . Is the new triangle also a right triangle ?

Solution
  Given BC = 12cm, AB = 5 cm, \angle B= 90^{\circ}
 

Steps of construction
1.   Draw a line BC = 12 cm
2.   From B draw AB = 5 cm which makes an angle of 90^{\circ} at B.
3.   Join AC
4.   Make an acute angle at B as < CBX
5.   On BX mark 3 point at equal distance at X_{1},X_{2},X_{3} .
6.   Join X_{3}C
7.   From X_{2}  draw X_{2}{C}'\parallel X_{3}C  intersect AB at {C}'
8.   From point {C}'  draw {C}'{A}'\parallel CA  intersect AB {A}'
Now  \bigtriangleup {A}'B{C}' is the required triangle and  \bigtriangleup {A}'B{C}'  is also a right triangle.

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